The equation of a circle $C$ is $x^2+y^2-2x-4y-31 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-2x) + (y^2-4y) = 31$ $(x^2-2x+1) + (y^2-4y+4) = 31 + 1 + 4$ $(x-1)^{2} + (y-2)^{2} = 36 = 6^2$ Thus, $(h, k) = (1, 2)$ and $r = 6$.